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I'm a newbie at sudoku. Why is the following 7 in Sudoku a mistake?

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  • $\begingroup$ @Duck I'm not seeing what is wrong with first row first column? $\endgroup$ – sshakir Aug 26 '18 at 18:44
  • $\begingroup$ There is nothing wrong it is just that you don't know which one it goes to. $\endgroup$ – Duck Aug 26 '18 at 18:47
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    $\begingroup$ If you still do not understand I am saying that you don't know what will be there later, so there could be a seven blocking where you put it when you finish the board $\endgroup$ – Duck Aug 26 '18 at 18:48
  • $\begingroup$ Also, if you want me to solve it I can but it is easier to just put the 7 one cell right of where you have it $\endgroup$ – Duck Aug 26 '18 at 18:52
  • $\begingroup$ Can you understand why you don't take the chance? If you finish everything else then you will realize that it will not go there $\endgroup$ – Duck Aug 26 '18 at 18:55
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Here is a logical reasoning for why that is not a seven, which should give you some insight into how to approach sudoku.

The board you are presented with is like this (and you clicked to place a 7 at position AZ as I have labelled it):

    A B C   D E F   G H I

Z   . . 5   3 6 .   4 . .
Y   9 6 2   . . 4   . 7 .
X   3 . 4   . 2 9   . 6 .

W   8 2 .   9 4 .   . 1 3
V   . 4 9   . 3 .   . 5 7
U   . . .   2 . .   9 8 .

T   4 . 6   . . 1   . . 2
S   . . .   6 9 3   . . 5
R   . . 3   . 8 .   . . . 

A cell cannot take a value if the value already appears in the same row, column, or 3-by-3-box.

Let's look at column C - 5,2,4,9,6, and 3 are already present, leaving 1,7, and 8 to be placed in some order into the empty cells at CW, CU, and CS.

Now lets look at row W - we can see that it contains a 1 (at HW) AND an 8 (at AW). This means CW (where row W and column C intersect) cannot be 5,2,4,9,6,3,1, or 8. This only leaves 7, so CW must be a 7:

    A B C   D E F   G H I

Z   . . 5   3 6 .   4 . .
Y   9 6 2   . . 4   . 7 .
X   3 . 4   . 2 9   . 6 .

W   8 2 7   9 4 .   . 1 3
V   . 4 9   . 3 .   . 5 7
U   . . .   2 . .   9 8 .

T   4 . 6   . . 1   . . 2
S   . . .   6 9 3   . . 5
R   . . 3   . 8 .   . . . 

Now CU and CS must be some arrangement of 1 and 8.

Looking at row U we can see an 8 (at HU) so CU must be a 1 (leaving CS to be an 8):

    A B C   D E F   G H I

Z   . . 5   3 6 .   4 . .
Y   9 6 2   . . 4   . 7 .
X   3 . 4   . 2 9   . 6 .

W   8 2 7   9 4 .   . 1 3
V   . 4 9   . 3 .   . 5 7
U   . . 1   2 . .   9 8 .

T   4 . 6   . . 1   . . 2
S   . . 8   6 9 3   . . 5
R   . . 3   . 8 .   . . . 

Now look at the 3-by-3-box around BV, it contains 8,2,7,4,9, and 1, leaving 3,5, and 6 for the other unassigned cells (AV, AU, and BU).

But row V contains two of these already - 3 at EV and 5 at HV, therefore AV, which is a cell in both the 3-by-3-box around BV and the row V can only be a 6:

    A B C   D E F   G H I

Z   . . 5   3 6 .   4 . .
Y   9 6 2   . . 4   . 7 .
X   3 . 4   . 2 9   . 6 .

W   8 2 7   9 4 .   . 1 3
V   6 4 9   . 3 .   . 5 7
U   . . 1   2 . .   9 8 .

T   4 . 6   . . 1   . . 2
S   . . 8   6 9 3   . . 5
R   . . 3   . 8 .   . . . 

Next we notice that since row V now has 6,4,9,3,5, and 7; column F has a 1 (at FT); and the central 3-by-3-box has a 2 (at DU) that the only number which may reside in FV is 8:

    A B C   D E F   G H I

Z   . . 5   3 6 .   4 . .
Y   9 6 2   . . 4   . 7 .
X   3 . 4   . 2 9   . 6 .

W   8 2 7   9 4 .   . 1 3
V   6 4 9   . 3 8   . 5 7
U   . . 1   2 . .   9 8 .

T   4 . 6   . . 1   . . 2
S   . . 8   6 9 3   . . 5
R   . . 3   . 8 .   . . . 

Similarly, column F now has 4,9,8,1, and 3; the 3-by-3-box around EY has a 2 (at EX) and a 6 (at EZ); and row Z has a 5 (at CZ) - therefore the only number which may reside in FZ is 7:

    A B C   D E F   G H I

Z   . . 5   3 6 7   4 . .
Y   9 6 2   . . 4   . 7 .
X   3 . 4   . 2 9   . 6 .

W   8 2 7   9 4 .   . 1 3
V   6 4 9   . 3 8   . 5 7
U   . . 1   2 . .   9 8 .

T   4 . 6   . . 1   . . 2
S   . . 8   6 9 3   . . 5
R   . . 3   . 8 .   . . . 

At last! We see that AZ cannot be a 7 (and neither can BZ) as there is already a 7 in row Z (at FZ).

Note that this also implies that there is a 7 at BX since one of the cells in the 3-by-3-box around BY must be a 7...

    A B C   D E F   G H I

Z   . . 5   3 6 7   4 . .
Y   9 6 2   . . 4   . 7 .
X   3 7 4   . 2 9   . 6 .

W   8 2 7   9 4 .   . 1 3
V   6 4 9   . 3 8   . 5 7
U   . . 1   2 . .   9 8 .

T   4 . 6   . . 1   . . 2
S   . . 8   6 9 3   . . 5
R   . . 3   . 8 .   . . . 

Now you may be able to apply similar reasoning to see that AZ should take a value of 1 and continue on to solve the puzzle!

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  • $\begingroup$ All great answers, but Jonathan, you went above and beyond to explain to me the thinking and strategy in detail. I really appreciate that. Thank you! $\endgroup$ – sshakir Aug 28 '18 at 22:57
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When you ask the puzzle to show you "mistakes" it doesn't just show you obvious things, like if you had put a 9 there when there's a 9 in that 3x3 already. It shows you anything that is different from the actual overall solution to the puzzle.

So when it says that 7 is a mistake, that means the actual completed solution to the sudoku doesn't have a 7 there. It doesn't mean "you should know that this is wrong from the information you can see now." There isn't enough information showing to be certain a 7 doesn't go there -- and there also isn't enough information to be certain that it does!

That's why many people turn off "show mistakes". I use it sometimes because I only record a number when I'm completely sure it must be right. For example the two in the top row has to be in the rightmost 3x3 and it can't be anywhere other than the middle cell of that top row of the 3x3. But the 7 you were placing, you can't be completely certain sure it goes there, you're just guessing. Just guessing and "show mistakes" don't go well together, at least not for me.

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  • $\begingroup$ Exactly what I was trying to say $\endgroup$ – Duck Aug 26 '18 at 21:19
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The seven can either go where you put it or right next to where you put it, there was a third chance that you would put it in the right place considering that I took no time to solve it so you just eliminated one out of the three possibilities so I recommend that you solve other parts first.

Here is the first box:

185 962 374

Also, you can't put it there because in the bottom left corner there would be a seven

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As @Kate Gregory pointed out, the 'hints' look at the final result. It technically is a mistake because you can't prove that 7 belongs there.

True sudoku is a game of logic not guess work with only one correct answer. You have to prove a number is eliminated from all other options.

The completed puzzle using a brute force. 7 goes between the 3 & 4.

185 367 429 962 514 378 374 829 561 ——————————— 827 945 613 649 138 257 531 276 984 ——————————— 496 751 832 218 693 745 753 482 196

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